# Newtonian Mechanics:friction, tension and acceleration

• Plebert
In summary, the problem involves three objects connected on a table with the coefficient of kinetic friction between one of the objects and the table being 0.350. The masses of the objects are m1 = 4.00 kg, m2 = 1.00 kg and m3 = 2.00 kg, and the pulleys are frictionless. The task is to draw free-body diagrams for each object, determine the acceleration and direction of each object, calculate the tensions in the two cords, and consider how the tensions would be affected if the tabletop were smooth. The relevant equations are F=ma, F(kinetic friction)=μkf x F(n), F(n)=m2xg, F(gravity)=mg
Plebert

## Homework Statement

Alright, the problem is as follows:Three objects are connected on a table. The coefficient of kinetic friction between the block of mass m2 and the table is 0.350.The other two masses are hanging perpendicular to the tabletop, suspended by wires on opposite sides of the table.From left to right, the objects have masses of m1 = 4.00 kg, m2 = 1.00 kg and m3 = 2.00 kg, and the pulleys are frictionless.

(a) Draw free‐body diagrams of each of the objects. (b)
Determine the acceleration of each object and their directions. (c) Determine the
tensions in the two cords. (d) If the tabletop were smooth, would the
tensions increase, decrease or stay the same?

## Homework Equations

These are the equations I've assumed to be relevant as none are given.
F=ma
F(kinetic friction)=μkf x F(n)
F(n)=m2xg
F(gravity)=mg
ƩF(m1)=((m1xg)-((μkf x(m2xg))+(m3xg)))

## The Attempt at a Solution

Well, for part (a) I can't really show you my free body diagrams, but I can tell you I derived this---->ƩF(m1)=((m1xg)-((μkf x(m2xg))+(m3xg)))
from drawing the forces associated with the mass.

When all relevant values are substituted in ended up with a net force of 14.985N In the positive "down" direction. Which when rearranged into the form

a=f/m
yields, a=3.7465m/s^2

for m1

I'm finding mass 2 much harder to model, I am not sure what role friction plays in defining the normal component of the reactionary force

for part c)
I'm assuming the tension in the wires is the Ʃf(x)
which would look something like, Ʃf(x)=(m1xg) + (m3xg) ?
I've never seen a relationship linking friction and tension...

so, is the tension independent of the friction? If so,this would mean the only forces causing tension are (m1xg) + (m3xg)

For part d) If I'm right in my above assertion, the tension would remain constant regardless of changes in μkf.

I know this question shouldn't be so complicated, but I'm just having trouble getting my head around everything.

Any help is priorly appreciated.

Plebert said:
Well, for part (a) I can't really show you my free body diagrams, but I can tell you I derived this---->ƩF(m1)=((m1xg)-((μkf x(m2xg))+(m3xg)))
from drawing the forces associated with the mass.
I don't understand how you got this from analyzing mass 1. Looks like you have forces acting on the other masses mixed in.

Start by identifying the forces acting on each mass.